Q. A coil of self inductance $50\ henry$ is joined to the terminals of a battery of e.m.f. $2\ volts$ through a resistance of $10\ ohm\ $and a steady current is flowing through the circuit. If the battery is now disconnected, the time in which the current will decay to 1/$e$ of its steady value is
A) $500\ seconds$
B) $50\ seconds$
C) $5\ seconds$
D) $0.5\ seconds$
Explanation
(c) Time in which the current will decay to $\frac{1}{e}$ of its steady value is $t = \tau = \frac{L}{R} = \frac{50}{10} = 5\ seconds$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.